Problem: Simplify the following expression: $y = \dfrac{7x^2- 13x- 2}{7x + 1}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(-2)} &=& -14 \\ {a} + {b} &=& &=& {-13} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-14$ and add them together. Remember, since $-14$ is negative, one of the factors must be negative. The factors that add up to ${-13}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${-14}$ $ \begin{eqnarray} {ab} &=& ({1})({-14}) &=& -14 \\ {a} + {b} &=& {1} + {-14} &=& -13 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 +{1}x) + ({-14}x {-2}) $ Factor out the common factors: $ x(7x + 1) - 2(7x + 1)$ Now factor out $(7x + 1)$ $ (7x + 1)(x - 2)$ The original expression can therefore be written: $ \dfrac{(7x + 1)(x - 2)}{7x + 1}$ We are dividing by $7x + 1$ , so $7x + 1 \neq 0$ Therefore, $x \neq -\frac{1}{7}$ This leaves us with $x - 2; x \neq -\frac{1}{7}$.